Chemistry 2000 (Spring 2006)
Problem Set #8: Chapter 18
Answers to Practice Problems
1.
Choose an appropriate acid, and describe how you would prepare 250 mL of a pH 3.50
buffer from a 1M solution of your acid and a 1M solution of NaOH.
Step 1: Choose an acid
The pKa of the acid should be as close to the target pH as possible (within 0.5 pH units).
The best acid in Table 17.3 is therefore nitrous acid (HNO2, Ka = 4.5 × 10-4, pKa = 3.35).
Step 2: Calculate the required ratio of acid to conjugate base (Henderson-Hasselbach equation!)
pH = pKa + log [conjugate base]
[conjugate acid]
[conjugate base] = 10pH-pKa = 10(3.50)-(3.35) = 1.4
[conjugate acid]
Step 3: Calculate the initial mole ratio for HNO2 and NaOH
Buffer requires 1.4 mol/L NO2- for every 1 mol/L HNO2.
Since the volume of the buffer is constant, it requires 1.4 mol NO2- for every 1 mol HNO2.
HNO2
+
OH→
NO2+
H2O
i
??
??
0 mol
c
e
1 mol
1.4 mol
Solving for ?? and ?? tells us that we will need 1.4 mol OH- for every 2.4 mol HNO2.
Step 4: Calculate the volume ratio for HNO2 and NaOH
Since both are added as 1 M solutions, the volume ratio is the same as the mole ratio:
M = n/V
therefore
V = n/M
VHNO2 : VOH = nHNO2 : nOH = 2.4 mol : 1.4 mol = 2.4 mL : 1.4 mL
1 mol/L 1 mol/L
MHNO2 MOH
Step 5: Calculate the volumes required for desired volume of buffer
Every 3.8 mL of solution was made by mixing 2.4 mL 1M HNO2 with 1.4 mL 1M NaOH.
VHNO2 = 250 mL buffer × 2.4 mL HNO2 = 1.6 × 102 mL HNO2
3.8 mL buffer
VNaOH = 250 mL buffer × 1.4 mL NaOH = 92 mL HNO2
3.8 mL buffer
Therefore, I would add 92 mL of 1M NaOH to 158 mL of 1M HNO2 to make a pH 3.50 buffer.
.. H
H ..
N
N
H
H
C
C
H
H
H
H
2.
(a)
Ethylenediamine (NH2CH2CH2NH2) is a base.
Draw the Lewis structure for ethylenediamine.
(b)
Is ethylenediamine monobasic, dibasic, or tribasic? (These are the equivalent terms to
monoprotic, diprotic, and triprotic.)
dibasic
(each nitrogen atom has a lone pair so each nitrogen atom can pick
up one H+)
(c)
Sketch a graph for the titration of a 0.01M ethylenediamine solution with 0.01 M HNO3.
Indicate the main species in solution (excluding water and spectator ions) in each region of
the titration curve.
3.
Zinc(II) hydroxide is added to pure water to make a saturated solution.
(a)
Calculate the pH of the solution.
Step 1: Set up ICE table
Zn(OH)2(s)
Zn2+(aq) + 2 OH-(aq)
Ksp = [Zn2+][OH-]2 = 3 x 10-17
i (M)
??
0
~0
c (M)
-x
+x
+ 2x
.
e (M)
?? – x
x
2x
Step 2: Solve for x.
Ksp = [Zn2+][OH-]2
3 × 10-17 = x (2x)2
4x3 = 3 × 10-17
x = [(3 × 10-17)/4]1/3 = 2 × 10-6
Check that we were safe to assume that initial
[OH-] would not affect equilibrium [OH-].
Since we only have 1 sig. fig., it doesn’t so
assumption was safe
Step 3: Calculate pH
[OH-] = 2x = 2(2 × 10-6 mol/L) = 4 × 10-6 mol/L
so
pOH = -log[OH-] = -log(4 × 10-6) = 5.4
so
pH = 14 – pOH = 14 – 5.4 = 8.6
(b)
Assuming that the solution is dilute enough that it has the same density as pure water,
calculate the equilibrium concentration of zinc in parts per million.
Step 1: Calculate equilibrium concentration of Zn2+(aq) in mol/L
[Zn2+] = x = 2 × 10-6 mol/L
Step 2: Convert concentration to ppm
density of water = 1 g/mL = 1000 g/L
[Zn2+] = 2 × 10-6 mol Zn2+ × 65.39 g Zn2+ × 1 L soln × 106 ppm = 0.1 ppm Zn2+
1L
1 mol Zn2+
1000 g soln
4.
Citric acid and phosphoric acid are both triprotic acids. Explain why it would be much more
difficult to calculate the pH of a 0.1 M solution of citric acid than to calculate the pH of a
0.1 M solution of phosphoric acid.
Generic deprotonation of a triprotic acid in three steps:
H3A + H2O
H2A- + HO-
Ka1 = [H2A-][HO-]/[H3A]
H2A- + H2O
HA2- + HO-
Ka2 = [HA2-][HO-]/[H2A-]
HA2- + H2O
A3- + HO-
Ka3 = [A3-][HO-]/[HA2-]
Phosphoric acid has Ka values that are separated by 5 orders of magnitude per step
(Ka1 = 7.5 × 10-3, Ka2 = 6.2 × 10-8, Ka3 = 3.6 × 10-13). As such, the successive
deprotonations of phosphoric acid can be treated as separate steps.
Citric acid has Ka values that are much closer together, separated by only 2 orders of
magnitude per step (Ka1 = 7.4 × 10-3, Ka2 = 1.7 × 10-5, Ka3 = 4.0 × 10-7). As such, the
successive deprotonations of citric acid cannot be treated separately. Consumption
of H2A- in the second step, for example, will be enough to affect production of it in
the first step. Thus, the three steps form one large, complex equilibrium (which is
more difficult to calculate than three smaller equilibria).
5.
Predict approximate values for K1 and K2 for telluric acid (H2TeO4). What type of acid is
H2TeO4?
H2TeO4 should behave similarly to H2SeO4 and H2SO4. Both have “very large”
values for Ka1 (i.e. Ka1 > 1) and Ka2 = 1.2 × 10-2. As such, I would predict that
H2TeO4 has Ka1 > 1 and Ka2 = 1.2 × 10-2.
Having Ka1 > 1 would make H2TeO4 a strong acid.